# How can this inverse of conditional statement be equivalent?

"A positive integer is a prime only if it has no divisors other than one and itself." The inverse of this conditional statement is : " A positive integer is not prime if it has divisors other than one and itself."
In logic the inverse isn't equivalent to the original statement. Can someone please clarify it for me??

• What is it you want clarified? – Prince M Feb 7 '16 at 11:29
• The inverse is not equivalent to the conditional but in your statement that is in fact the definition of prime so it is a biconditional and if the converse is true then the inverse is true. – Prince M Feb 7 '16 at 11:33

"A positive integer is a prime if and only if it has no divisors other than one and itself." "A positive integer is not prime if and only if it has divisors other than one and itself."

These are equivalent in logic.

The statement you gave is equivalently "If a positive integer is prime, then it has no divisors other than one and itself."

The converse is "If a positive integer is composite, then it has a divisor which is neither itself nor one."

However, in definitions, we often sloppily say "if" when we mean "if and only if". For example, you might see "a metric space is compact if every open cover has a finite subcover" as a definition of "compact", when really it is intended to be read as "a metric space is compact if and only if every open cover has a finite subcover". In the context of a definition, "if" is usually meant as "iff".

Therefore, in this instance, the converse is equivalent to the original statement, because we are working with a definition.

• I believe the converse you give is actually the inverse. If a conditional is $P \implies Q$ the inverse is $\neg P \implies \neg Q$. – Prince M Feb 7 '16 at 11:37
• @PrinceM Ah, I had never heard that terminology before. Since the inverse is logically equivalent to the converse, I deem it not to matter :) I'll delete the "you mean converse" line. – Patrick Stevens Feb 7 '16 at 11:38
• The converse would be $Q \implies P$ or "if a positive integer has a divisor which is neither itself nor one then the positive integer is composite" and it would matter except in this statement it is a biconditional so of course the converse is true thus the inverse is also true! – Prince M Feb 7 '16 at 11:43
• Thanks Patrick Stevens & Prince M. – Nadim Baraky Feb 9 '16 at 17:39

Let $n$ a positive integer, $p$ "$n$ is a prime.", and $q$ "$n$ has no divisors other than one and itself." then first sentence can be written as $$p \rightarrow q,$$ and second sentence can be written as $$\neg q\rightarrow \neg p.$$ They are equivalent, because $\neg q\rightarrow \neg p$ is a contrapositive of $p \rightarrow q$, not an inverse. If you want to suggest inverse, then $$\neg p \to \neg q,$$ equivalent to "If a positive integer is not prime it has divisors other than one and itself.", or "A positive integer has divisors other than one and itself if it is not prime."

• The title of the askers question is propositional logic, not predicate logic. Propositional logic does not include quantifiers. – Prince M Feb 7 '16 at 11:40
• The point of my post is "because $\neg q(n) \rightarrow \neg p(n)$ is a ...", not that I used quantifiers. – choco_addicted Feb 7 '16 at 11:46
• Then why did you bother quantifying it? And the asker is asking about the inverse not the contrapositive of the conditional. In this case the inverse is true because the affirmative is biconditional so the converse must be true and the inverse is the contrapositive of the converse so it must also be true. – Prince M Feb 7 '16 at 11:51
• How you quantified his second statement is not correct. He clearly has it written as $\neg p(n) \implies \neg q(n)$ – Prince M Feb 7 '16 at 11:55
• Second statement is "A positive integer is not prime if it has divisors other than one and itself.", equivalent to "If a positive integer has divisors other than one and itself it is not prime." Then $\neg q(n) \rightarrow \neg p(n)$ is correct. – choco_addicted Feb 7 '16 at 12:00

I believe you are going wrong with words. From Rosen's Discrete Mathematics, $p \rightarrow q$, can be expressed this way:

• $p$ only if $q$.

• $q$ if $p$.

It means that if your first statement is $p \rightarrow q$, your second statement is actually $\neg q\rightarrow \neg p$ (contrapositive), not $\neg p\rightarrow \neg q$ (inverse).

• Yeah I was confused at first. Thanks anyways :) – Nadim Baraky Feb 9 '16 at 17:40